So in any book that discusses the Dirac equation in Relativistic Quantum Mechanics (For example you can check Bjorken and Drell, Greiner, or Ashok Das), whenever they take the dagger of the Dirac eq., they replace $i \frac{\partial}{\partial x}$ by $-i \frac{\partial}{\partial x}$, which is the same as $\hat{p} \rightarrow \hat{p}^\dagger=-\hat{p}$. In quantum mechanics, however, we found that $\frac{\partial}{\partial x}^\dagger=-\frac{\partial}{\partial x}$ which came from doing integration by parts and the vanishing of the boundary term. How do we reconcile both views and is $\hat{p}$ really anti-Hermitian in RQM?

  • $\begingroup$ $\hat{p}^\dagger=-\hat{p}$ is wrong. Correct is $\hat{p}^\dagger=\hat{p}$, meaning $\hat{p}$ is Hermitian, like all observables. From $\hat{p} = - i \frac{\partial}{\partial x}$ it follows $\hat{p}^\dagger = - i \frac{\partial}{\partial x}$. So there is no contradiction. $\endgroup$ – Thomas Fritsch May 13 at 9:29
  • $\begingroup$ But $\hat{p} = i \partial$ and as I said they write $(i \partial )^\dagger = -i\partial$ so $\hat{p}^\dagger=-\hat{p}$ $\endgroup$ – Ammar Amgad May 13 at 9:32
  • 1
    $\begingroup$ You've lost a minus sign: $(i \partial )^\dagger = i^* \partial^\dagger = (-i)(-\partial) = i \partial$ $\endgroup$ – Thomas Fritsch May 13 at 9:37
  • $\begingroup$ Yes, that's exactly what I am talking about! If you check Ashok Das Lectures on QFT, or Bjorken and Drell RQM, you will find that when taking the dagger of the Dirac eq., they replace $\partial$ by $\partial ^\dagger = \partial$ not $-\partial$. So this is my question, why? $\endgroup$ – Ammar Amgad May 13 at 9:44

$\hat{p}^\dagger = -\hat{p}$ is not correct. Momentum operator churns out the momentum of a particle/system which is a real observable. Hence, it has to be Hermitian, since the expectation value of any anti-Hermitian operator must be purely imaginary and hence can't be observed.

How do we see momentum is Hermitian:

  • $\dfrac{d}{dx}$ is anti-Hermitian: $\left(\dfrac{d}{dx}\right)^\dagger = -\dfrac{d}{dx}$.

  • Since $\dfrac{d}{dx}$ is anti-Hermitian, $\left(i\dfrac{d}{dx}\right)$ must be Hermitian: $\left(i\dfrac{d}{dx}\right)^\dagger= - i\left(-\dfrac{d}{dx}\right) = \left(i\dfrac{d}{dx}\right)$.

  • From this relation the momentum operator, $\hat{p} = - i \hbar\dfrac{d}{dx}$ is indeed Hermitian: $\hat{p}^\dagger=\hat{p}$.

I guess you are trying to say this probably: $\hat{p}^* = \left(-i \hbar\dfrac{d}{dx}\right)^* = \left(i\hbar\dfrac{d}{dx}\right) = -\hat{p}$.

  • $\begingroup$ No, I mean dagger. To understand why I am asking this, check Ashok Das Lectures on QFT, p.44, eq.2.73. $\endgroup$ – Ammar Amgad May 13 at 9:47
  • $\begingroup$ @AmmarAmgad It is not the equation for momentum, right? It is the Dirac equation: $i \frac{\partial}{\partial t} = H \psi$ and not $i \hbar \frac{\partial }{\partial x} $ $\endgroup$ – exp ikx May 13 at 10:00
  • $\begingroup$ Yes, where $H=\mathbf{\alpha}.\mathbf{p} + \beta m = -i\mathbf{\alpha}.\mathbf{\nabla} + \beta m$ $\endgroup$ – Ammar Amgad May 13 at 10:06
  • $\begingroup$ When getting the dagger, he writes $H^\dagger=i\mathbf{\alpha}.\mathbf{\nabla} + \beta m$ with a left arrow over the nabla. $\endgroup$ – Ammar Amgad May 13 at 10:08
  • $\begingroup$ I don't understand why $(-i\nabla)^\dagger$ should equal $i\nabla$, can you elaborate? $\endgroup$ – Ammar Amgad May 13 at 10:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.